关于android手机图案锁屏的计数

无心人

2=>1,3,4,5,6,7,9
3=>2,4,5,6,8
4=>1,2,3,5,7,8,9
5=>1,2,3,4,6,7,8,9

无心人

6=>1,2,3,5,7,8,9
7=>2,4,5,6,8
8=>1,3,4,5,6,7,9
9=>2,4,5,6,8

wayne

不错，这种方法好是好，只是有一种情况没考虑吧。
比如，前面如果已经经过5的话，后面4可以直接跳到6的

参考6#可行方案的2516图
路径{8,5,7,4,6}

无心人

待我写个程序来

无心人

1. 
   
2. #include <stdio.h>
   
3. 
   
4. int mask[10] = {1,0,0,0,0,0,0,0,0,0};
   
5. int next[10][10] =
   
6. {0,0,0,0,0,0,0,0,0,0,
   
7. 5,2,4,5,6,8,0,0,0,0,
   
8. 7,1,3,4,5,6,7,9,0,0,
   
9. 5,2,4,5,6,8,0,0,0,0,
   
10. 7,1,2,3,5,7,8,9,0,0,
    
11. 8,1,2,3,4,6,7,8,9,0,
    
12. 7,1,2,3,5,7,8,9,0,0,
    
13. 5,2,4,5,6,8,0,0,0,0,
    
14. 7,1,3,4,5,6,7,9,0,0,
    
15. 5,2,4,5,6,8,0,0,0,0,
    
16. };
    
17. int counter[10] = {0,0,0,0,0,0,0,0,0,0};
    
18. int graph[10] = {0,0,0,0,0,0,0,0,0,0};
    
19. int nextNumber = 0;
    
20. int graphPostion = 0;
    
21. 
    
22. void first( void )
    
23. {
    
24.   int i;
    
25.   for (i = 1; i <= 9 ; i ++)
    
26.   {
    
27.     mask[i] = 1;
    
28.         graph[graphPostion] = i;
    
29.         circle( graphPostion + 1 );
    
30.     mask[i] = 0;
    
31.   }
    
32. }
    
33. 
    
34. void circle( int graphPostion )
    
35. {
    
36.   int i;
    
37.   int noPostion = 1;
    
38.   int n = graph[graphPostion - 1];
    
39.   for (i = 1; i <= next[n][0]; i ++)
    
40.   {
    
41.     if (mask[next[n][i]] == 0)
    
42.         {
    
43.           noPostion = 0;
    
44.           mask[next[n][i]] = 1;
    
45.             graph[graphPostion] = next[n][i];
    
46.           counter[graphPostion + 1] ++;
    
47.           circle( graphPostion + 1 );
    
48.           mask[next[n][i]] = 0;          
    
49.         }
    
50.   }   
    
51.   
    
52.   if (noPostion)
    
53.   {
    
54.     printf("find: ");
    
55.     for (i = 0; i < graphPostion; i ++)
    
56.     {
    
57.           printf("%1d  ", graph[i]);
    
58.     }
    
59.         printf("\n");
    
60.   }       
    
61. }
    
62. 
    
63. int main ( void )
    
64. {
    
65.   int i;
    
66.   first( );
    
67.   for (i = 4; i <= 9; i ++)
    
68.     printf("%d: %d\n", i, counter[i]);
    
69.   return 0;
    
70. }
    

复制代码

4: 1400
5: 5328
6: 16032
7: 35328
8: 49536
9: 32256

wayne

1. PointsToBeChecked[{{a_,b_},{c_,d_}}]:=Module[{g=GCD[c-a,d-b]},If[g>1,{a+# (c-a)/g,b+# (d-b)/g}&/@Range[g-1],{}]];
   
2. n=3;ddd=Flatten[Table[{{i,j}},{i,0,n-1},{j,0,n-1}],1];
   
3. Table[{ii,Product[n^2-k,{k,0,ii-1}],ddd=Select[Flatten[Table[Join[ii,List@#]&/@Complement[Flatten[Array[List,{n,n}]-1,1],ii],{ii,ddd}],1],Length[PointsToBeChecked[#[[-2;;-1]]]]==0&];Length[ddd]},{ii,2,n^2}]

复制代码

对算法做了改进，反而跟无心人的答案对上了。
{2, 72, 56},
{3, 504, 304},
{4, 3024, 1400},
{5, 15120, 5328},
{6, 60480, 16032},
{7, 181440, 35328},
{8, 362880, 49536},
{9, 362880, 32256}}

wayne

算得4*4的情况：
2,240,172
3,3360,1696
4,43680,15580
5,524160,132264
6,5765760,1029232
7,57657600,7286016
后面就算不动了

无心人

证明俺算法没错了哦

mathematica

取出所有的排列,然后一个一个地判定!

gogdizzy

看了下无心人的程序，能计算#6给出的合法情况吗？
5->2->7->3->4

因为5是已经标记过的，所以可以从7直接到3了。